It is a well-known task to answer whether a node is an ancestor to another one with time complexity O(n) for preprocessing and O(1) for a query.
Each time you enter the node in recursive function DFS, you increase the time counter by one and call it an "entrance" time, and after recursively processing its sons, you increase the time counter by one and call it an "exit" time. Here is the pseudocode.
DFS(u, p): // u is the current node, p is its parent
increase the timer counter
in[u] = timer
for all (u, v) in edges:
if v != p:
DFS(v, u)
increase the timer counter
out[u] = timer
Now u is an ancestor of v 
.
Is there someway to implement the same algorithm but with non-recursive DFS?
Of course. You can mantain stack with pairs (vertex, flag), where flag = [in, out]. So, while stack not empty, you pick up pair from vertex. If it's "in" pair, you push (vertex, out) to stack and push childs (child_vertex, in) after. If it's "out" pair, you simply update out array.
Instead of pairs, you can mantain in/out information in array, cause first visit to vertex is always in and second visit is always out.